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Consider Maxwell's equation (without source):

$$ \partial_\mu F^{\mu \nu} = 0 \implies \partial_\mu \partial^\mu A^\nu = \partial_\mu \partial^\nu A^\mu.$$

Can we find a pair of classical field configurations $A^\mu(x),A'^\mu(x)$ such that they both satisfy the equation above (assuming similar boundary conditions) but are not related to each other by a gauge transformation of the type:

$$A'^\mu(x) = A^\mu(x)+\partial^\mu \varphi(x) \quad ?$$

If it's impossible, how could we argue/show this?

Answer: Thanks to my2cts' answer, I've found 2 solutions $A$ and $A'$ not related by a gauge transformation : $A^\mu=(0,e^{-i(t-y)},0,0)$ and $A'^\mu=(0,0,e^{-i(t-x)},0)$. It makes sense since they both give rise to different EM fields, which are invariant under gauge transformations.

Jasmeru
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    I don't think one can answer this question until the boundary conditions are defined. – akhmeteli Feb 14 '19 at 05:52
  • I think that reading the comments and answer to my question posed in the following link answer this question. https://physics.stackexchange.com/q/452308/ – InertialObserver Feb 14 '19 at 07:20
  • @akhmeteli I completely agree and rarely if one ever sees them mentioned in the context of the vector potential but why? – hyportnex Feb 14 '19 at 12:54
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    Just for the record: there an even more trivial example. $A=0$ is a solution, the set of all solutions that differ from it by a gauge transformation are $A=\mathrm{d}\phi$. All such solutions have $F=0$. Clearly there exist solutions with non zero $F$. – MannyC Feb 14 '19 at 18:09

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Of course there are many solutions that are unrelated by a gauge transformation, namely plane waves of any frequency, propagation direction, polarisation and phase and superposition thereof.

my2cts
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  • @my2cts If I understand your point, the idea is that we can find 2 linearly independent solutions $E,B$ and $E',B'$ that solve Maxwell's equation, and since the EM fields are unaffected by a gauge transformation we cannot relate them to each other? – Jasmeru Feb 14 '19 at 15:24
  • It is not obvious that all plane waves satisfy the same boundary condition, as I wrote in my comment, this depends on the boundary condition. – akhmeteli Feb 16 '19 at 03:22